If A is diagonalizable, is $p(A)$ diagonalizable?

Question

Problem: Let $A$ any $n\times n$ diagonalizable matrix. Is $p(A)$ diagonalizable for any polynomial p(x)?

My attempt: I set $$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$ I know that a matrix is diagonalizable iff there are n linearly independent eigenvectors.
So I set $Ax=qx$ (q is eigenvalue), and after some substituting, i got $$q(A)x=q(p)x$$ This is the part where I got stuck. I get a feeling that I should divide cases when A has n linearly independent eigenvectors by having n distinct eigenvalues, or if it has an eigenvalue with geometric multiplicity larger than $1$… Any hint please?

Answer 1

If $M$ is such that $M^{-1}AM$ is equal to some diagonal matrix $D$, then $p(M^{-1}AM)=M^{-1}p(A)M$, which is a sum of diagonal matrices. More precisely, if$$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0,$$then$$p(M^{-1}AM)=a_nD^n+a_{n-1}D^{n-1}+\cdots+a_1D+a_0\operatorname{Id}.$$Therefore, $p(M^{-1}AM)$ is a diagonal matrix too.